Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <algorithm>
#include <iostream>
#include <vector>
#include <math.h>
#include <set>
#include <map>
#include <string>
#include <stack>
#include <queue>
#include <iomanip>
#include <numeric>
#include <tuple>
#include <bitset>
#include <complex>
#include <unistd.h>
#include <cassert>
#include <cctype>
#include <random>
#include <time.h>
#define _USE_MATH_DEFINES
#define _GLIBCXX_DEBUG
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> plglg;
typedef pair<double, ll> pdlg;
typedef tuple<int, int, int> tiii;
typedef tuple<ll, ll, ll> tlglglg;
typedef tuple<double, double, double> tddd;
typedef complex<double> xy_t;
typedef vector<ll> vll;
typedef vector< vector<ll> > matrix;
#define REP(i, x, y) for(ll i = (ll)x; i < (ll)y; i++)
#define DREP(i, x, y, d) for(ll i = (ll)x; i < (ll)y; i += (ll)d)
#define PER(i, x, y) for(ll i = (ll)x; i > (ll)y; i--)
#define DPER(i, x, y, d) for(ll i = (ll)x; i > (ll)y; i -= (ll)d)
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
double pi = 3.141592653589793;
ll mod = 1000000007;
int intmax = 2147483647;
int intmin = -2147483648;
ll llmax = 9223372036854775807;
ll llmin = -9223372036854775807;
int iinf = intmax / 8;
ll inf = llmax / 8;
double eps = 1e-12;
void yesno(bool ok, string yes = "Yes", string no = "No") {
    if (ok) {
        cout << yes << endl;
    } else {
        cout << no << endl;
    }
}

struct edge {
    ll to, cost;
};

int V;
vector<edge> G[1000000];
vector<ll> d1, d2;

void addedge(int st, int ed, ll co) {
    edge e;
    e.to = ed;
    e.cost = co;
    G[st].push_back(e);
}

void dijkstra(int s, vector<ll> &distance) {
    priority_queue<plglg, vector<plglg>, greater<plglg> > que;
    distance = vector<ll>(V, inf);
    distance[s] = 0;
    que.push(plglg(0, s));

    while (!que.empty()) {
        plglg p = que.top();
        que.pop();
        ll v = p.second;
        if (distance[v] < p.first) {
            continue;
        }
        int len = G[v].size();
        for (int i = 0; i < len; i++) {
            edge e = G[v][i];
            if (distance[e.to] > distance[v] + e.cost) {
                distance[e.to] = distance[v] + e.cost;
                que.push(plglg(distance[e.to], e.to));
            }
        }
    }
}

ll snum[200010];
ll tnum[200010];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N, M, s, t;
    cin >> N >> M >> s >> t;
    V = N;
    s--;
    t--;
    REP(i, 0, M) {
        ll a, b;
        cin >> a >> b;
        a--;
        b--;
        addedge(a, b, 1);
        addedge(b, a, 1);
    }
    dijkstra(s, d1);
    dijkstra(t, d2);
    REP(i, 0, N) {
        if (d1[i] != inf) {
            snum[d1[i]]++;
        }
        if (d2[i] != inf) {
            tnum[d2[i]]++;
        }
    }
    ll ans = 0;
    REP(i, 0, d1[t] - 1) {
        ans += snum[i] * tnum[d1[t] - i - 2];
    }
    cout << ans << endl;
}